For customers purchasing a refrigerator at a certain appliancestore, let A be the event that the refrigerator wasmanufactured in the U. S., B be the event that therefrigerator had an icemaker, and C be the event that thecustomer purchased an extended warranty. Relevant probabilities arebelow.

(a) Construct a tree diagram consisting of first-, second-, andthird-generation branches and place an event label and appropriateprobability next to each branch.

(b) Compute P(A intersectB intersectC).

(c) Compute P(B intersectC).
2

(d) Compute P(C).
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(e) Compute P(A | B intersectC), theprobability of a U. S. purchase given that an icemaker and extendedwarranty are also purchased.

P(A) = 0.73 P(B | A) = 0.90 P(B | A' ) =0.76
P(C | A intersectB) =0.84 P(C | A intersectB' ) =0.65
P(C | A' intersectB) =0.71 P(C | A' intersectB' ) =0.32

Part a : The tree diagram is attached with the solution.

Part b: The probability of P(A∩B∩C) is 0.5519.

Part c: The probability of P(B∩C) is 0.6976.

Part d: The probability of P(C) is 0.7658.

Part e: The probability of P(A∣B∩C) is 0.7658.

Explanation:

The solution is as follows.

Data Given is as

P(A) = 0.73

P(B | A) = 0.90

P(B | A' ) =0.76

P(C | A∩B) =0.84

P(C | A∩B' ) =0.65

P(C | A' ∩B) =0.71

P(C | A' ∩B' ) =0.32

Part a:

The remaining values of Probabilities for the tree formation are calculated as follows

P(A')=1-P(A)=1-0.73=0.27

P(B'|A)=1-P(B|A)=1-0.90=0.10

P(B' | A' )=1-P(B | A' ) =1-0.76=0.24

P(C' | A∩B) =1-P(C | A∩B)=1-0.84=0.16

P(C' | A∩B' ) =1-P(C | A∩B' ) =1-0.65=0.35

P(C' | A' ∩B) =1-P(C | A' ∩B) =1-0.71=0.29

P(C' | A' ∩B' ) =1-P(C | A' ∩B' )=1-0.32=0.68

so the probability tree is given as attached in the diagram.

Part b:

P(A∩B∩C)=P(A)P(B∣A)P(C∣A∩B)

From the above data in part a

P(A)=0.73, P(B|A)=0.90 and P(C∣A∩B)=0.84

                =0.73×0.90×0.84

               =0.55188

               ≈0.5519

The probability of P(A∩B∩C) is 0.5519.

Part c

P(B∩C)=P(A∩B∩C)+P(A'∩B∩C)

           = [P(A)P(B∣A)P(C∣A∩B)+{1−P(A)}P(B∣A′)P(C∣A′∩B)]

From the above data

P(A)=0.73,P(B∣A)=0.9,P(C∣A∩B)=0.84,P(B∣A′)=0.76 and P(C∣A′∩B)=0.71

P(B∩C)= [P(A)P(B∣A)P(C∣A∩B)+{1−P(A)}P(B∣A′)P(C∣A′∩B)]

           =(0.73×0.9×0.84)+(1−0.73)×0.76×0.71

           =0.55188+0.145692

           =0.697572

           ≈0.6976

The probability of P(B∩C) is 0.6976.

Part d

P(C)=P(A∩B∩C)+P(A′∩B∩C)+P(A∩B ′∩C)+P(A′∩B ′∩C)

      =P(A)P(B∣A)P(C∣A∩B)+[1−P(A)]P(B∣A′)P(C∣A′∩B)+P(A)[1−P(B∣A)]P(C∣A∩B′)+            

        [1−P(A)][1−P(B∣A′)]P(C∣A′∩B′)

Here from above data

P(A)=0.73,P(B∣A)=0.9,P(C∣A∩B)=0.84,P(B∣A′)=0.76,P(C∣A∩B′)=0.65,P(C∣A′∩B′)=0.32 and P(C∣A′∩B)=0.71

P(C)=P(A)P(B∣A)P(C∣A∩B)+[1−P(A)]P(B∣A′)P(C∣A′∩B)+P(A)[1−P(B∣A)]P(C∣A∩B′)+            

        [1−P(A)][1−P(B∣A′)]P(C∣A′∩B′)

     ={0.7x0.9x0.84}+{(1-0.73)x0.76×0.71}+{0.73x(1-0.9)x0.32}+{(1-0.73)x(1-          

       0.76)x0.32}

     =0.55188+0.145692+0.04745+0.020736

     =0.765758

     ≈0.7658

The probability of P(C) is 0.7658.

Part e

From part (b),P(A∩B∩C)=0.55188

From part (c), P(B∩C)=0.697572

By  conditional probability rule,

P(A∣B∩C)= P(A∩B∩C)/P(B∩C)

       = 0.55188/0.697572

              =0.791144

             ≈0.7911

The probability of P(A∣B∩C) is 0.7658.

answer: 2 weeks hbu

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